$\newcommand{\Spec}{\operatorname{Spec}}$ Cross-post from Math.SE, hopefully people more knowledgeable in the field will see the question here on MO.

It is a well-known fact that a smooth projective variety over $\mathbb Q$ has good reduction almost everywhere, i.e. everywhere apart from finitely many points. In the language of schemes, this is almost obvious -- given projective equations of the variety $X/\mathbb Q$, for some $n$ we can construct a projective model $\mathcal X\to\Spec\mathbb Z[1/n]$ of $X$. Since a projective morphism is proper and the singular locus of $\mathcal X$ is closed, the set of points of $\Spec\mathbb Z[1/n]$ with singular fibers is closed, hence finite (since the generic fiber is smooth).

When I've first learned this proof, it seemed somewhat clear to me this works for smooth *proper* varieties over $\mathbb Q$, but my professor has pointed out that while the latter part of the argument goes through, there may be problems in constructing a model over a subscheme of $\Spec\mathbb Z$. He couldn't, off the top of his head, answer whether such varieties could have infinitely many places of bad reduction, but he speculated the answer is yes. This is my question:

Can a smooth proper variety over $\mathbb Q$ have infinitely many places of bad reduction?

I have speculated about a much stronger failure as well:

Can a smooth proper variety over $\mathbb Q$ have bad reduction

everywhere(i.e. at every finite prime)?

Since every proper curve is projective, the example necessarily has dimension at least $2$.

Note: I have added a reference-request tag because I strongly suspect the answer has been already addressed in the literature, but my searches were unsuccessful, mostly returning results about varieties with everywhere good reduction.